Question

Mark the correct alternative in the following question:

A box contains 3 orange balls, 3 green balls and 2 blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing 2 green balls and one blue ball is

$\left(\mathrm{a}\right)\frac{167}{168}\left(\mathrm{b}\right)\frac{1}{28}\left(\mathrm{c}\right)\frac{2}{21}\left(\mathrm{d}\right)\frac{3}{28}$

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ O\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{of}\mathrm{drawing}\mathrm{a}\mathrm{orange}\mathrm{ball}, \\ G\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{of}\mathrm{drawing}\mathrm{a}\mathrm{green}\mathrm{ball}\mathrm{and} \\ B\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{of}\mathrm{drawing}\mathrm{a}\mathrm{blue}\mathrm{ball} \\ \mathrm{We}\mathrm{have}, \\ \mathrm{n}\left(O\right)=3,\mathrm{n}\left(G\right)=3\mathrm{and}\mathrm{n}\left(B\right)=2 \\ \mathrm{Also},\mathrm{total}\mathrm{balls}=8 \\ \mathrm{Now}, \\ \mathrm{P}\left(\mathrm{drawing}2\mathrm{green}\mathrm{balls}\mathrm{and}\mathrm{one}\mathrm{blue}\mathrm{ball}\right)=\mathrm{P}\left(GGB\right)+\mathrm{P}\left(GBG\right)+\mathrm{P}\left(BGG\right) \\ =\mathrm{P}\left(G\right)\times \mathrm{P}\left(G|G\right)\times \mathrm{P}\left(B|GG\right)+\mathrm{P}\left(G\right)\times \mathrm{P}\left(B|G\right)\times \mathrm{P}\left(G|GB\right)+\mathrm{P}\left(B\right)\times \mathrm{P}\left(G|B\right)\times \mathrm{P}\left(G|BG\right) \\ =\frac{3}{8}\times \frac{2}{7}\times \frac{2}{6}+\frac{3}{8}\times \frac{2}{7}\times \frac{2}{6}+\frac{2}{8}\times \frac{3}{7}\times \frac{2}{6} \\ =\frac{1}{28}+\frac{1}{28}+\frac{1}{28} \\ =\frac{3}{28} \end{gathered}$$

Hence, the correct alternative is option (d).